Question

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is $V_{max}$. Then the average speed of the particle during the SHM is :

• A. $\frac{V_m}{\pi }$
• B. $\frac{V_m}{2\pi }$
• C. $\frac{2V_m}{\pi }$
• D. $\frac{3V_m}{\pi }$

Answer 1

Answer: (C)

$Vm=A\omega$
$Vavg=\frac{distancetraveled}{timetaken}$
$=\frac{A}{T}$
$T=\frac{2\pi }{\omega }$
$Vavg=\frac{4A\times \omega }{2\pi }=\frac{2A\times \omega }{\pi }$

$Vavg=\frac{2V_m}{\pi }$